text:
How much force can the landing gear of a Piper Arrow withstand?
I'm currently working on my CFI-A in a Piper Arrow and i was wondering if anyone knows how much force the landing gear can withstand? The reason I ask is because I have never been asked this question and I know there has to be a concrete answer out there. For example, aircraft in the Navy that need to land on a aircraft carrier have to go through a series of landing gear stress tests.

Its not a requirement for GA aircraft to test landing gear systems to identify the point of failure. It would be difficult anyway, there are a lot of factors that go into it. Age of the aircraft, corrosion, previous hard landings, etc. There is no magic number here. Navy aircraft have that because they have highly specialized landing systems which are maintained to a high standard.

The Arrow is certificated under 14 CFR 23, and the landing gear loading limits are specified in the ground loads and landing gear sections. I don't know much about this subject and the regulations are fairly technical, but 23.473 apparently defines the load limit. I don't really understand it myself, but it looks like the landing gear has to support at least 2.67 times the aircraft's weight in a landing:

No inertia load factor used for design purposes may be less than 2.67,
  nor may the limit ground reaction load factor be less than 2.0 at
  design maximum weight, unless these lower values will not be exceeded
  in taxiing at speeds up to takeoff speed over terrain as rough as that
  expected in service.

In reality they should be designed to withstand more than this, as there are manufacturing uncertainties which can affect a component's performance.
Someone who's actually an engineer could probably explain it better (or correct me, as the case may be!).
As for testing, my limited understanding of the certification process is that the manufacturer has to prove to the FAA that the requirements in part 23 have been met, but there's no specific way of doing that. And indeed, 23.721(c) says:

Compliance with the provisions of this section [i.e. the landing gear section] may be shown by
  analysis or tests, or both.

Thank you so much for this response this was very helpful!.


text:
Is this the relationship between rpm and thrust of a Turbofan?
In this graph from the FAA Airplane Handbook Chapter 15, page 5:

It looks like it is talking about "jet engines" but is it engine type specific or it is for both turbojet/fan/prop/shaft? It looks right about the more rpm the engine is running, the more thrust it would produce, but is the relationship shown here for jet engine or piston engine?

Does this answer your question? What is the relationship between thrust and engine RPM in a turbojet?

The basic way jet engine and piston engine produce thrust on an airplane is the same. They accelerate a given amount of air which in reaction provide a forward force called thrust. The faster the fan or the propeller spins, the more air it can accelerate thus increasing thrust. There is a good answer there. The reason why they don't operate at the same percentage of available RPM is efficiency. It might be true that the thrust increase with RPM for any aircraft engine, but propeller have huge losses at high speed the propulsive efficiency going down dramatically. And this is mostly due to the tip vortex appearing at the tip of each blade as you can see in the picture below. As turbofan are enclosed in a casing, those tip vertices don't develop and you have less losses.

Now to your question, as you can see the curve start a 50% of the maximal RPM with a thrust rating very low. This is typically the case for turbofan where the fan design has been optimised for higher RPM. At low RPM the AOA ouf the blade is not optimal and the efficiency is quite bad.

That is a generic graph for jet engines where most of the thrust is achieved in the high rpm regime. The meaning is simple. Thrust is a function of the amount of air the engine can ingest. Or if you wish, the number of molecules. This in turn relates to the amount of rpms and the density of the air: Thrust = rpm x density. The more rpms the more particles ingested, the denser the air the more particles as well. In a practical example: When climbing with constant thrust, the jet engine should increase the rpms to compensate the less air density. Same thing will happen when temperature increases. See below a table with N1 rpms for the CF34-10 jet engine. In order to maintain constant thrust the FADEC will adjust the rpms.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Does a failure to respond to FAA questions on ME disqualify for sport pilot?
I had submitted a medical evaluation to pursue a private pilot certification. The FAA had requested further evaluation by specialists. I didn't respond to their questions. Does this disqualify me for a sport pilot license?

Yes I believe so.
While the Sport Pilot License only requires that you are medically qualified to get a driver's license there is a stipulation that if you had been denied an FAA medical previously then you cannot just use your driver's license as your medical.  If you don't respond to their questions then you will most certainly be denied the medical.
At this point I think it is best to answer their questions and finish out the process.  If your medical condition is under control and manageable the FAA will usually eventually grant you the medical however, a lot of paperwork, time and money may be required.
This is why you want to do your homework before the medical and be sure that you can pass it.  If you don't pass the medical then you cannot just simply fall back to the Sport Pilot License.

"there is a stipulation that if you had been denied an FAA medical previously then you cannot just use your driver's license as your medical" - I'm not sure that's correct. Under 14 CFR 61.23(c)(2)(ii), if you're a sport pilot using your driver's license to satisfy the medical requirements, then that privilege goes away when you apply for a medical, not when you're denied a medical. That privilege is then restored if you are issued a medical (and that privilege does not expire when the medical expires). I could be wrong, but that's my reading of the law.

@TannerSwett:  That makes sense.  When I said denied I meant that they applied but did not get the medical.  I think the OP thinks that if he doesn't respond to the FAA's questions he would not get denied and could still get the Sport's license but that is indeed not correct.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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If the altitude of the aircraft is given by the pitot tube, in icing condition with heat on, how is real altitude /speed interpretation done?
If the altitude  of the aircraft is given by the pitot tube, in icing condition with pitot heat on how is  real altitude /speed interpretation done? Because the air density and temperature is not realistic, due to the fact that pitot tube is hot .

The pitot tube is used for airspeed, not altitude measurement.

No, because I asked about altitude not relative speed. WHAT instrument do that?

@GeorgeGeo You asked about a heated pitot tube, which is answered in the linked duplicate. Altitude is derived from static pressure, which is usually measured by a static port, which is not heated. The two functions can be combined (see What is the difference between a pitot tube and a pitot-static tube?), but that does not change the validity of the answer in the linked duplicate.

The pitot tube measures pressure, not density. The pressure is not affected by the heating of the tube.

Would it be affected by viscosity? And would viscosity be affected by temperature?

@Bianfable: I cannot quantity the effect, so my question, it should be more important at low velocity since the inertia effect will be lower (the linked article is not accessible).

@mins Interesting, thanks for the link. It's probably a small effect (since viscosity is increasing less than linear with temperature), but it might make for an interesting new question.


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How are the landing lights of the legacy F-18 Hornet seen before landing on the carrier?
The LSO will be able to see the nose gear landing lights to make sure that the aircraft's  angle of attack is at $8.1^\circ$ if the aircraft is "on speed". Are these lights (all 3 of them) attached to the landing gear on all the time or at their  desired level of angle of attack?

The lights match the standard AoA indexer in the cockpit.

When the airplane is on AoA, the center amber light is illuminated, as is the amber ‘donut’ on the AoA display.  If the airplane is at a higher AoA, the upper green light illuminates as does the green caret on the AoA display.  Shallow AoA, the lower red light and red caret lights.

Just curious, is the main bright white light not illuminated during the carrier approach and landing?  If it was, seems those colored lights would be hard to see--.  Also do those colored lights have a more specific and correct name than "landing lights"?

I’m not sure on that one. It may be that landing lights are not used during carrier flight operations for that reason.

Correct, landing lights are always off for carrier ops, it's in the checklist to verify.  P.S.  Good answer, but I don't think "no" at the beginning is necessary or correct.  The question is either/or...

The colored approach lights that the LSOs reference are not unique to the legacy Hornet, every carrier based aircraft is similarly equipped.  These lights come on as soon as the gear is down and locked, and as such can provide secondary external confirmation of landing gear status in case there is an unsafe indication in the cockpit.  During carrier operations the bright white landing light is left off to ensure that they can be seen by the LSOs.


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How is it possible for the MIG 21 to have full rudder to the left, but the nose wheel move freely to the right and then straight or to the left?
How is it possible that the MIG 21 has the rudder to the left, but the nose wheel moves freely to the right, and then straight or to the left on the runway?
See timestamp 1 min 11 sec in this video:

Davidw mentioned the book "Red Eagles".  It has a passage about steering the MiG-17 and says the MiG-21 used exactly the same system.  In the following passage, he is quoting pilot Jose Oberle.
Red Eagles: America's Secret MiGs, Steve Davies, 2011

Taxiing the airplanes was a bit of a trick.  If you wanted to turn
right, you pushed the left [sic][a] rudder bar in and you pulsed this lever on
the control stick.  That dumped the pneumatic pressure to the brake
on the left wheel and transferred it to the right wheel.   You would
get the free-swinging nose wheel to start to turn to the right, and
then as you got ready to come out of that turn you'd have to push the
left rudder bar past neutral and you'd have to start pumping the
little paddle again to get the power back on the left wheel, to get
the nose to straighten out so you could taxi straight!

Per this, the nosewheel is not part of the steering system.
a: most definitely a typo, see for instance the following pilot's experience.
I Was Lucky: I Got to Be a Pilot, Paul "PK" Kimminau, 2017, p. 161

After checking the systems out, I taxied to the runway. That was an
experience in itself!! On US aircraft, the brakes are on the rudder
pedals, that is, you push on the tops which would rotate forward and
apply braking. Also on US aircraft, if you wanted to turn, you pushed
the rudder pedal just as if you were using rudder to fly the aircraft
and power steering would turn the nose wheel. On the MiG-17, the
braking was done with a lever on the control stick. You squeezed it
and applied brakes. If you wanted to turn, you pushed the rudder pedal
in and squeezed the brake lever, and it would apply braking to that
wheel so you turned in that direction.

The directions don't check out—it says to turn right you first push left rudder and then to straighten out again you push the left rudder again?

@JanHudec You push the left rudder and dump the pressure.

@Sneftel and then you push left again, but pump the pressure the other way? Is that how you understand it?

@JanHudec Essentially, yeah. I don't fully grok the description, but it seems like you're using the pneumatic pump as a pressure release the first time, and as an actual pump the second time.

You got the directions wrong. If you want to turn right, you push the RIGHT pedal and depress the brake lever. This will apply right braking through a brake differential valve and make the nose wheel go right as it is free castoring. In the video the pilot applied full left rudder or pedal to straighten the nose after the right turn.

Please don't think they are my words.  As cited in the answer, the quote is from pilot Jose Oberle, as recorded in the Davies book.  Since the question is about the nosewheel, I don't think it matters.  I quoted the section to give context, but the main fact from there is that the nose is described as free-swinging.  If a part of the description were flipped, that shouldn't matter much.

Most larger jet aircraft do not have a full time link between rudder pedals and nosewheel steering.  In planes I have flown there is a nosewheel steering button on the control stick.  It is spring loaded, and as long as it is held down (i.e. during taxi and initial take-off roll) the nosewheel will steer with the rudder pedals.
This is done because the steering sensitivity needed to make a 90 degree turn taxiing at low speed would be way too much on the runway at high speed.  For smaller GA aircraft this isn't so much of an issue, and the relative simplicity of not needing a method of disconnecting the two overrides any small advantage.

So, presumably, at the moment in question in the video, the button is not pressed, and the nosewheel is in free-swivel mode, and the turn is being accomplished with differential braking?  (Answer might benefit from including some of this-- or providing alternate explanation if this is wrong.  Is there a nosewheel steering "tiller"?)

@Quiet Flyer, why are you presuming that the NWS button is not pressed? And I agree that it could have a free castering nosewheel, controlled by differential braking, but I have not flown a Mig-21 so I can't confirm. And typically tillers are only used on much larger aircraft.

Due to the rudder and nosewheel being deflected in opposite directions, as per the original question--

OK, I see.  Well, I won't try to guess what kind of system it has.  I think my answer might be useful for someone with only Cessna 172 type knowledge and experience.  I will let it ride for now and let the votes play out...  If anyone severely trumps me with a credible answer I can always delete it later.

I know that the Mig 21's have a bicycle hand brake on their stick, but I guess this is the symetrical brake sistem (maybe in conjunction with the rudder pedal pushing or more importantly  singled use to brake straight line )But if you push left rudder then the deflection is left as the direction of the aircraft, don't you think?

I try to limit my speculating to topics that I am confident have some similarity with things I am familiar, with or have knowledge of.  The Soviets did some very different things with aircraft design so your guess is as good as mine.

It might very well be a free castering nosewheel with no steering function at all, and that's the assumption I made watching the video (all the Grumman general aviation airplanes are like that).   Even the B-29 had a free castering non-steerable nosewheel, making it quite a challenge to maneuver at low speed.  Also the Martin B-26 IIRC.

From reading the book Red Eagles, about the 4477th TES, which flew MiG's in the 1980s, I recall that the steering system on the MiG-17, and possibly others, was considered unusual and difficult for the USAF pilots to learn.


text:
With a separation of 1000 feet, in flight is there any danger of severe wake turbulence?
Air Canada Flight 190 at 36600 experienced severe wake turbulence due to a Boeing 747-400 United 896 Flight at 37000. Assume the worst possible combination of civilian aircraft under standard speeds and air conditions, if they are separated by 1000 can the smaller plane in rear still suffer severe wake turbulence and at what sort of distance separation?

Summary: 1000 ft vertical separation is not enough for wake turbulence from a heavier aircraft flying above to decay. The wake turbulence will however be significantly less strong by then. A lateral distance of 25 NM is required to avoid all wake turbulence.

If both aircraft are level, the situation looks something like this:

En-route, the vortices evolves in altitudes at which the rate of decay leads to a typical persistence
of 2-3 minutes (min), with a typical sink rate of about 400 ft/min. Wakes will also be transported
by wind.
Considering the high operating air speeds in cruise and the standard 1000 ft vertical separation in
Reduced Vertical Separation Minima airspace, wake can be encountered up to 25 nautical miles
(NM) behind the generating aeroplane. The most significant encounters are reported within a
distance of 15 NM.

(EASA Safety Information Bulletin 2017-10R1)
If the wake sinks at 400 ft/min, it will reach the lower altitude 2½ minutes later. The typical persistence time is listed as 2-3 minutes, so a strong wake still exists, although at lower intensity.
For a large difference in aircraft weight, this can still result in severe turbulence. An example is the incident involving a Bombardier Challenger CL604 and an Airbus A380 in 2017 (as mentioned by Digital Dracula in the comments):

The analysis of the flight data of both aircraft showed that at 0838:07 UTC the A380
had passed the CL604 overhead with a vertical distance of 1,000 ft. [...]
Within one second the bank angle increased to 42° to the right. At
the same time the aileron deflection to the left increased to 20° and the vertical
acceleration to 1.6 g. In the following second vertical acceleration changed to -3.2 g. [...]
At the time of the accident [the flight attendant]
had been standing in the middle of the cabin preparing the service. Four of the six
passengers had also not been seated. In her recollection the airplane had turned
three times around its longitudinal axis, during which the occupants had been thrown
against the ceiling and the seats. Several of the passengers suffered injuries, some
of which were bleeding.

(German Bundesstelle für Flugunfalluntersuchung: Interim Report)

The situation is more dangerous, if one or both of the aircraft are climbing or descending. This can result in a situation, where the aircraft are much closer vertically than 1000 ft:

In the en-route phase of flight, three major factors contribute to increase the likelihood of wake
turbulence:

Crossing traffic situation: If crossing traffic is climbing or descending in proximity (either the
generating or following aeroplane), the wake generating aeroplane might cross the follower’s
trajectory with minimum time for decay, so stronger wake turbulence might be encountered.

Thermal tropopause altitude: Wake vortex decays more slowly below the tropopause, if there
is an increased risk of encountering severe wake turbulence.

Weight of the generating aeroplane: Heavier aeroplane types generate stronger wake vortices
and are likely to induce more severe wake turbulence encounters, especially for smaller
aeroplane types.

(EASA Safety Information Bulletin 2017-10R1, emphasis mine)
This exact situation existed for the Air Canada 190 incident you mentioned:

It was established that the aircraft had been climbing in accordance with its ATC clearance from FL350 to cruise altitude FL370 behind a United Airlines operated Boeing 747-400 en route from Hong Kong to Chicago and level at FL 370 and that the incident had occurred as the A319 had passed FL366 at a distance of 10.7 nm in trail behind the Boeing.

(SKYbrary: A319 / B744, en-route near Oroville WA USA, 2008)
The vertical separation was only 400 ft and at a distance of 10.7 NM the wake turbulence had not decayed much yet.

Bombardier Challenger CL604 and an Airbus A380 in 2017- so lucky the Bombardier pilot knew what he experienced and responded appropriately.


text:
How does the maneuverability of the FJ-2 Fury and F-86 compare?
The North American FJ-2 Fury was the US Navy version of the F-86 Sabre jet.
I want to know if G maneuver limits were in favor of the F-86 or the FJ-2 Fury.

What exactly do you want to know? The Wikipedia article explains the differences quite well.

I want to know  G  maneouvres   limits were in favor of the F86 OR   FJ2 Fury

OK, I updated your question to include that. Feel free to edit again, if you want to know something else.

Interesting. Apparently it can sometimes happen on ASE, with some browsers, that one can be viewing the older version of a question, and can submit an upvote or downvote, not knowing that the question has actually been changed. Then when one refreshes the question and sees the change, it is already too late to submit a new vote. Good to know.


text:
May the tower use the runway while an emergency aircraft is inbound?
I have a question regarding an aircraft in emergency. Of course it will have  priority for landing but if the position of this aircraft is very far, like for example 20 minutes or more away,
would the tower be allowed to clear other aircraft to land or take off? Is it permissible for the tower to do so?
Is there any law or statement in the civil aviation regulation about such things?

since you are asking about laws, could you please specify the jurisdiction?

No. The tower is immobile and cannot access the runway.

@Davidw http://www.militarysystems-tech.com/files/militarysystems/imagecache/gallery_main/supplier_images/Mobile%20Elevating%20Towers-1-l.jpg

FAA Order JO 7110.65W - ATC Procedures, Chapter 10 outlines how ATC should handle procedures for emergency situations
Right at he very beginning, we see the most important part:

10-1-1(d)
  Because of the infinite variety of possible emergency situations, specific procedures cannot be prescribed. However, when you believe an emergency exists or is imminent, select and pursue a course of action which appears to be most appropriate under the circumstances and which most nearly conforms to the instructions in this manual.

In other words, they don't dictate what controllers should do in an emergency; they are to use their best judgment about how to handle it.
So, a controller is free to continue using a runway for both arrivals and departures if they believe doing so won't interfere with the emergency aircraft.
In fact, for non-urgent emergencies like gear extension failures, I've heard them hold the emergency aircraft to get as many others in and out as possible before the runway is closed for a gear-up landing.

Wouldn't that be a good idea anyway to reduce the amount of fuel the aircraft is carrying, which reduces the weight (I assume that can only help) and the fire risk?

@gnasher729 if there are parallel runways they will also asked if the emergency aircraft will use the outer of the two to so they can keep the inner one open or get it opened earlier. This helps prevent another plane from needed to declare an emergency for fuel reasons.

@gnasher729 I normally take off with full tanks, even for a short hop, and I’m not holding for up to 6 hours to burn off excess fuel. Though I will try to crash somewhere with ARFF available for exactly that reason.

Yes, generally speaking, the runway can be used. That said, whether or not it will actually be done depends on many different factors, such as the type of emergency, the size of the airport (maybe there are multiple runways), the current traffic load, the emergency response required etc. The fire marshall would most likely be the only person who could actually decide to close the runway, but ATC could, in practise, decide to not let anyone else land until the emergency is over. It depends.
From an ATC point of view, it may make more sense to get other flights on the ground before the emergency is due to land, instead of having to deal with airborne holdings and possible diversions. However, the risk of another aircraft breaking down and blocking the runway is non-zero, so if the inbound emergency has absolutely no other choice than to land on that specific runway, maybe just closing it for other traffic is the better option.


text:
What happens to a plane that goes to / from an enclave when the surrounding country closes its airspace?
When Pakistan closed its airspace, planes that had routes flying over Pakistan had to deviate around that country.
What would happen to planes en route to or from Lesotho, if South Africa closed its airspace?
Are the planes simply unable to leave or enter the country, or there is a law / regulation about this situation?

South Africa has an agreement with Lesotho and Swaziland for single point of management of the upper airspace of the three countries. Lesotho and Swaziland retain control over lower airspace. For example, Swaziland has a rule that nothing may fly over 50 ft over the ground without permission. A detective was once arrested and fined for flying a drone and taking pictures above that altitude. Swaziland also has a law that bans witches broomsticks from flying above 150 meters above the ground. No, it's not a joke.
Because of that agreement, it would be highly unlikely that South Africa would close airspace in such a way as to deny entry and exit to aircraft flying to and from Lesotho and Swaziland. I believe that the agreement might even require all three governments to agree before closing of the airspace, and in case of lack of agreement there would have to be corridors in place for aircraft to navigate to/from Lesotho and Swaziland.
That said, enforcement of the agreements may be another story. Paper and reality often diverge. Given the overwhelming disparity in economic, military and regulatory resources, I am not sure Lesotho and Swaziland would be able to enforce continued access if South Africa were to close its airspace for whatever reason.

Good to know, from that article, that Swazi witches don't use brooms for transport, only to fling potions.

it would be no different than in case say China and Russia both closing their airspace for aircraft flying into and out of Mongolia. There'd be no way for aircraft to enter or leave that country.
Nothing special about being an enclave there.

I hardly see how it answers the question. The question is "what happens [in this case]?"

This answer simply points out that the situation isn't unique.  If any landlocked country or state was denied access to the surrounding territory it would be locked in.  The portion of the question about whether there is a law or regulation to prevent it in this specific case is legit, otherwise "what would happen" is pretty open ended.


text:
Does center of gravity change between ramp weight and takeoff weight?
Most documentation gives percentage weight distrubtion between main and nose gear for ramp weight. I'm curious if this weight distribution would be the same for takeoff weight.

Is %MAC a relationship between the nose gear and the main gear for ramp weight?  Perhaps... to determine maximum force exerted by each wheel.  For takeoff, enroute and landing operations, most airplanes use a relationship between the leading edge and trailing edge of the wings.  Thus 0% MAC puts the CG at the leading edge of the wing.  Personally, can't imagine why they wouldn't use the same formula for ramp weight.  It would provide the details necessary without extra pilot workload.

The difference would be from fuel burned during taxi, which is generally a very small fraction of the total fuel, so the impact on CG would be tiny. And, since burning fuel tends to bring the CG more into, rather than out of, limits, the "who cares" impact is essentially nil.

So burning fuel moves the center of gravity forwards, towards the nose?

Depends where the CG is in relation to the center of the fuel tank being burned (which might not be in the same exact place depending on which tank is being burned right now). I'm not certain that there won't be exceptions to a particular generalization in that regard. For the 737-800, as best I recall, adding fuel to the wing tanks moves the CG in one direction, then adding fuel to the fuselage tank (filled last, burned first) moves it in the other.

Swept wing a/c with center tanks, the CG moves aft as the center is burned, then back forward as the wings are burned.  With no center tanks, it just moves forward.  On a straight wing a/c with wing tanks, the shift is negligible, very slightly forward as the fuel's CG in the wing box is a bit aft of the plane CG.

Thanks guys, awesome responses. During taxi, do you know which tank is being used, so as to determine which direction the center of gravity is likely going in?

On a 737 taxiing out for takeoff, if the total fuel load is more than about 18,200 lbs (full wings + 1,000 or more in the center tank), you're feeding from the center tank. Less than that, you're feeding from the mains.

Just for the heck of it, I ran a weight and balance for a 747-400 with a ramp fuel load of 200000 lbs, a taxi fuel burn of 2000 lbs, and an otherwise empty airplane. It moved the ramp CG from 27.9% MAC to 27.8% MAC. That equates to 0.3278 inches.

@Terry  You guys had a tank in the horizontal stab, yes? Was fuel burned from there, or just transferred?   ((Good to hear from you - was thinking you might have thoughts on this thread.))

@Terry you're back! Good to see you!

@RalphJ I flew the -100/200s so no stab tanks. The only -400s I have data for had no stab tanks, either because they never had them (they were optional I think) or because they were taken out when converted to freighters. So unfortunately I'm unable to be of help.

@FreeMan I've never left. I check here almost daily, but as time wears on the answers I can give with any certainly diminish. Plus dementia is playing havoc with my brain. LOL

Well, good to see you chiming in, @Terry. Always interesting to hear your insights. And the offer to be your grandson (complete with great- and great-great-grandkids) still stands! Love hearing your stories!!

Great answer above. Acceleration also plays a negligible task regarding this as fractions of fuel are forced to settle backward in each fuel tank; this somethimes causes more noticeable spills under the wing of some overfilled airliners only when they start to roll for take off - as fuel is forced backwards and follows the trailing edge swept angle outwards until they reach the vents... and leaks.


text:
How can my CD0 estimation be improved?
In this question (link) @PeterKämpf asked me to give a basis for my results. I will share my method for estimation here to see if you guys agree it is correct and how I can improve on it. Note that I use ISA equations where relevant.
Firstly, I estimate the Breguet constant from the PL-R diagrams of the A319-100 and B737-700 (found in the airport planning manuals). For that I use this equation, which is essentially the Breguet equation applied between points A and B (A being MPL+MTOW, B is MTOW+MFW):
$$ K_{sem,i} = \left( \frac{R_B - R_A}{\ln \left( \frac{OEW + PL_A}{OWE + PL_B} \right)} \right)_{sem,i} $$
From the PL-R diagrams: A319-100: R_A = 1463 km, PL_A = 15825 kg, R_B = 5800 km, PL_B = 3800 kg, OEW = 41100 kg (weight variant 00). B737-700: R_A = 3956 km, PL_A = 16860kg, R_B = 6236 km, PL_B = 10800 km, OEW = 38342 kg. — Results: K_A319 = 18277 km, K_B737 = 19607 km.
Using the definition of the Breguet constant, I can estimate their average aerodynamic efficiency, with the following equation:
$$ \left( \frac{L}{D} \right)_{sem,i} = K_{sem,i} \left( \frac{gc_{j,cr}}{V} \right) $$
For that I need cruising speed (A319: M0.78 at 11900m ISA = 230.15 m/s, B737: M0.785 at 11705m ISA+10ºC = 236.91 m/s) and sfc in cruising conditions. I have sfc at 35000 ft and M0.8 16.98 mg/Ns for the A319 and 17.02 mg/Ns in the same conditions for the B737). I adjust the sfc to cruising conditions using the following proportionality (note that beta = 0.5 for turbofan engines):
$$ c_j = \text{cte}(M)^\beta \sqrt{T} $$
I get adjusted values of 16.68 mg/Ns for the A319, 17.2 mg/Ns for the B737. Average aerodynamic efficiencies come out at 13 and 13.96, respectively. I can calculate the average lift coefficient using L=W:
$$ C_{L_{sem,i}} = \left( \frac{2 \frac{W_{cr}}{S_W}}{\rho V^2} \right)_{sem,i} = \left( \frac{2 \cdot 0.8 \frac{W_{TO}}{S_W}}{\rho V^2} \right)_{sem,i} $$
where I've estimated average cruising wing loading as 80% of take-off wing loading. The take-off wing loading of the A319 is 5129 Pa and for the B77 it's 5485 Pa. Density is in cruising conditions. Average lift coefficients come out as 0.4906 and 0.5023, respectively. Using the definition of the polar curve, I calculate CD0 using the following expression:
$$ C_{D0_{sem,i}} = C_{L_{sem,i}} \left( \frac{L}{D} \right)^{-1}_{sem,i} - \frac{C^2_{L_{sem,i}}}{\pi A_{sem,i} \varphi_{sem,i}} $$
where I've taken $\varphi = 0.9$ for both planes (since they have winglets). The aspect ratio of the A319 is 9.5, 9.4 for the B737. CD0 comes out as 0.0274 for the A319 and 0.0265 for the B737. This is off somewhat from the 0.014-0.02 range given by Torenbeek for high-subsonic jets.
Excuse the length, but hopefully you guys can help me improve this estimation or detect errors.

Please don't use images for formulas. I converted them to MathJax. If I made a mistake somehwere, please edit again (here is a MathJax tutorial and reference).

Does this method only take cruise conditions into account? Meaning: how is fuel consumption until reaching cruise altitude accounted for?

The other method for estimating the $C_{D0}$ is given in Torenbeek 5.3.1 and Appendix F:  aeroplane component drag is compared with the friction drag of an equivalent flat plate having the same wetted area and length. $C_D$ of wing, fuselage, tailplane and engine nacelles are determined and  added up.

Thank you for sharing your calculation! Now I can see how you arrived at the results. Never mind the length: All that detail is essential for improving your results.
Regarding the Breguet equation I prefer to use the one here and here. This, using your data for the A319, gives
$$\frac{L}{D} = \frac{g\cdot b_f\cdot R}{v\cdot ln\left(\frac{m_2}{m_1}\right)} = 17.63$$
with $v$ = 230 m/s, $R$ = 4,337,000 m, $g$ = 9.81 m/s², $b_f$ = 0.00001668 kg/Ns, $m_1$ = 62,925 kg and $m_2$ = 74,950 kg. You will notice that I changed the masses: You cannot use two different payloads for the Breguet equation; instead, all mass differences must be from fuel consumption. I now used the difference in payloads to compute the mass ratio between range 1 and range 2 and use that for the mass of the fuel consumed to cover $R$.
The result is a bit lower than what is usually used for the A319, which probably results from different payloads used for the two different ranges.
Your lift coefficient looks quite credible, even though you could do better than pick a mass reduction of 80% once you know the mass together with the altitude at one point during cruise. The Breguet equation is only valid under the assumption of a constant lift coefficient which is only possible when the airplane climbs continuously as fuel is consumed. In reality, flights get assigned flight levels and do a step climb if traffic control permits. But using the lift coefficient from one data point should be close enough.
Also your way of calculating $c_{D0}$ is good in principle, although I would prefer to lower the wing efficency to 0.85 – swept wings rarely perform better, winglets notwithstanding. If you allow me to use the more usual $\epsilon$ instead of $\varphi$, and if L/D is indeed 18 for the A319, the zero-lift drag coefficient becomes
$$c_{D0} = \frac{c_L}{\frac{L}{D}} - \frac{c_L^2}{\pi\cdot AR\cdot\epsilon} =  \frac{0.5}{18} - \frac{0.25}{3.14159\cdot9.5\cdot0.85} = 0.0179$$
which lies within the range given by Torenbeek. I only did the numbers for the A319; the B737 data should produce a very similar result.
Your biggest error was in applying the Breguet equation with the wrong masses. Also the masses themselves are too low: Just OEW and payload will not allow the airplane to fly: You need to add some fuel. And the mass fraction must be from the same flight at different times, with the range the distance covered between those times.

EDIT: Now that I know where your payload-range numbers are from, I can check your results. Since the inclination of the part of the diagram where payload is traded against fuel mass is the same for all MTOW figures, the resulting L/D numbers should also be the same.
From the brown line (75,500 kg MTOW) I read 7,500 km range with 10,000 kg payload and 5,000 km with 16,000 kg payload. This means the airplane consumes 6,000 kg of fuel for 2,500 km range. Since the range/fuel ratio becomes worse with increasing range, this figure cannot be used in isolation but helps to figure out the mass at the end of a median trip. Let's use a 6,250 km flight which is the middle of that range where payload is traded against fuel mass, so a fuel mass of 15,000 kg is consumed at the end of the flight. That makes the mass ratio 75,500 / 60,500 = 1.248 and the L/D = 17.6 with your SFC number. Close enough to 18, given my rounding of the mass figures.

Thanks very much for the reply Peter, it is very detailed. However I think perhaps I didn't express myself correctly. The first equation is Breguet applied at point A of the PL-R diagram, point B of the PL-R diagram, subtract the equations and you get that expression. MTOW cancels out (that's where the fuel weight would be). Let me know if you understood it like that, I think perhaps your comment on applying the wrong masses is based on a misunderstanding. If I have misunderstood, ¿could you elaborate further on this point? Thanks in advance.

To clarify my comment, I mean that these are two different flights, but the Breguet constant (K) is always the same for any conceivable flight. So both flights take off at MTOW and their landing weights are OEW + PL_A and OEW + PL_B.

@stevederekson555 MTOW or fuel mass cannot cancel out in a fraction where numerator and denominator are sums. Also, landing weight should include fuel reserves. Again, you cannot take the masses from two different flights.

I'm using this form of the Breguet equation: R = K*Ln(TOW/LW). It's a natural logarithm, so the subtractions turn into divisions. You're right that I'd have to include reserves though, that's a good point. But given that K is always the same for the aircraft I don't see why you can't perform this operation.

@stevederekson555 That is not how it is presented in the question. The fraction TOW/LW is correct, but must be from the same flight. TOW is never given in the question.

It is given in the question, at point A defined at MTOW+MPL and point B defined at MTOW+MFW. It's MTOW/LW_1 and MTOW/LW2. So you're left with two different Breguet equations. You have everything except K, so you can obtain the expression for K from this, because K should be the same for both equations.

@stevederekson555 MTOW is maximum take-off mass. There cannot be a flight mass that is MTOW plus payload. Payload is part of MTOW. Please edit your question and fill in the numbers for each range separately. I think your original mistake is to assume that the take-off weight cancels out because it is the same in both equations. But that is misapplying the Breguet equation.

Sure but it's TOW/LW. So if it takes off at MTOW, that includes the payload and the fuel weight. And in both flight conditions it takes off at MTOW, which is the same value for both flight conditions. So basically it's this: On the PL-R diagram you obtain two points, each has a payload, a range and a takeoff weight (namely MTOW). You can apply the Breguet equation Range = K*Ln(MTOW/LW). The LW is OEW + payload at that point (add reserves if needed). So you're left with a 2x2 system where everything is known but K, which is one and only for the aircraft. Why wouldn't MTOW cancel?

I really don't think it's a misapplication of the Breguet equation, I don't see how it can be.

Note that when I say MTOW+MPL I mean it takes off at MTOW with maximum payload, so this payload is included within MTOW. It's just a point on the PL-R diagram.

@stevederekson555 Look at each point separately. Yes, both should give the same K, but do them one after the other. What L/D do you get each time? It should be also the same and there is no need to solve a 2x2 system. Next you tell me the L/Ds also cancel out.

Right so if you get the same K you should get the same L/D if the plane flies in the same cruising conditions (using design values). So when I calculate L/D like this from K (using the definition of K) for the A319, I get a value of 13, very different to what you get (16.7). Also Roskam gives typical values for cruising L/D for high subsonic jets are in the range of 13-15. I'm rereading your answer to see why but I can't find the issue.

@stevederekson555 https://aviation.stackexchange.com/questions/1738/how-does-the-boeing-787-dreamliners-glide-ratio-compare-to-other-airliners/13975#13975

This gets to the core of the question. Try to obtain K from the PL-R diagram for the A319, then L/D from K using the method we discussed here. If you get a lower than expected result, why do you think it is? I guess that is what I was trying to get at with this question.

The PL-R diagram can be found here on page 129, I obtained L/D = 13 from the 64000kg MTOW weight variant: https://www.airbus.com/aircraft/support-services/airport-operations-and-technical-data/aircraft-characteristics.html

@stevederekson555 Thank you a lot for the link! Please see my use of its figures which gives an L/D of 19 rsp. 17.6 in the added part of the answer. You must leave a sufficient fuel reserve and cannot land with dry tanks!

Solid answer. Thanks.


text:
What light signals or visual markings indicate that turns should be made to the right at class G airports?
FAR 91.126 states:

§91.126 Operating on or in the vicinity of an airport in Class G airspace.
(a) General. Unless otherwise authorized or required, each person operating an aircraft on or in the vicinity of an airport in a Class G airspace area must comply with the requirements of this section.
(b) Direction of turns. When approaching to land at an airport without an operating control tower in Class G airspace—
(1) Each pilot of an airplane must make all turns of that airplane to the left unless the airport displays approved light signals or visual markings indicating that turns should be made to the right, in which case the pilot must make all turns to the right; and
[...]

Are there examples of "approved light signals or visual markings indicating that turns should be made to the right"?

Here's the example from the FAA's Airplane Flying Handbook (chapter 7):

If there are no other aircraft present, the pilot should check
  traffic indicators on the ground and wind indicators to determine
  which runway and traffic pattern direction to use. [Figure 7-2] Many
  airports have L-shaped traffic pattern indicators displayed with a
  segmented circle adjacent to the runway. The short member of the L
  shows the direction in which the traffic pattern turns are made when
  using the runway parallel to the long member. The pilot should check
  the indicators from a distance or altitude well away from any other
  airplanes that may be flying in the traffic pattern.

Here's what it looks like at Huntsville Executive (KMDQ), which has a right-hand pattern for runway 36:

4.2.6 Right-hand traffic
When displayed in a signal area, or horizontally at the end of the runway or strip in use, a right-hand arrow of conspicuous colour (Figure A1-9) indicates that turns are to be made to the right before landing and after take-off.

  Figure A1-9

Source: ICAO Annex 2, Rules of the Air

The "L"-shaped marks near the tetrahedron or wind tee — adjoining the "segmented circle" — show the traffic pattern direction.
Intended to be viewed from above, not from ground level.

http://www.flightlearnings.com/2011/12/16/wind-direction-indicators/
The figure above indicates "left traffic" (left turns) when taking off or landing to the north, right traffic when taking off or landing to the south, left traffic when taking off or landing to the east, and right traffic when taking off or landing to the west.  Typically we'd see this situation if there were some noise-sensitive area, obstacle, etc southwest of the airport.
Or as depicted in the FAA's "Airplane Flying Handbook" as noted in another answer:

This indication is a little more confusing than the one in the first diagram.  To interpret this indication we can imagine that there is a miniature runway running through the tetrahedron or wind sock.  The key point is that the "L" shape we need to pay attention to is the one at the approach end of that imaginary miniature runway, not the departure end, in relation to our planned take-off or landing.  For aircraft taking off or landing to the northeast, "left traffic" is indicated by the "L" shape on the lower right of the circle.  For aircraft taking off or landing to the southwest, "left traffic" is indicated by the "L" shape on the upper right of the circle.


text:
Is flying direct to a navaid considered a leg with track guidance?
When I design SID or Missed approach procedure with a typical sequence of initial legs like "Climb to 5000 feet on R-123 ABC VOR/DME, turn left/right direct to ABC VOR/DME", the regulation checker of my flight procedure design software (IDS FPDAM) often warn about the below PANS-OPS standard because it considers the leg turning direct to the VOR/DME without track guidance:

3.1.2 Track guidance shall be provided:
b) within 10.0 km (5.4 NM) after completion of turns for turning departures.

Then most of the time I have to assign a specific course to the navaid (VOR Radial or NDB Bearing) for the aircraft to intercept on the way back to the navaid.
I see that if a specific course is not required, the aircraft will turn direct to the navaid on any course which is suitable with its performance. But the aircraft always have track guidance regarding the navaid.
Then, is the software's warning about the leg turning direct to a navaid without track guidance correct or not?

I see that if a specific course is not required, the aircraft will turn direct to the navaid on any course which is suitable with its performance. But the aircraft always have track guidance regarding the navaid.

In the earlier paragraph you stated:  "...most of the time I have to assign a specific course..."
Is it fair to presume that you only get the warning when you do NOT assign a specific course?
I am unfamiliar with the software you are using, but as you pointed out aircraft performance, (speed, angle of bank, etc.) as well as wind, will cause any turn direct to a navaid to vary, perhaps significantly.  If you do not specify a radial or ground track to intercept the pilot would track whatever course they rolled out on, or could simply home in on a curving path.
By definition then; if you don't specify a track there can be no track guidance because there is no track to provide guidance to, right?
That being the case, your sentence in bold is incorrect, and the answer to your title question is NO, flying direct to a navaid (without a track specified) is not considered a leg with track guidance.
I think that is what the software warning is trying to tell you.  Does that make sense?

An FMS wants to see a sequence of magenta lines from point to point to point. If the procedure is to overfly AAAAA and then turn toward BBB and where-ever that turn finishes (uncertain due to unknown winds & TAS & etc), go from there to BBB, then that clean sequence of points & legs between them isn't defined at the outset. While various FMC's can do various things (like RF legs), it sounds like the rules the OP has wants "and then intercept course ### to BBB" as the solution there. (I'd expect defining AAAAA as fly-past rather than fly-over might work too.)

The problem with “turn direct” is that depending on the aircraft speed and the wind speed/direction, the ground track may be completely different.
This appears to violate the rule requiring “track guidance”, which from your problem I assume means defining a single fixed track over the ground for every flight, regardless of speed or wind. That is presumably to limit the terrain/obstacle clearance surface to something manageable.
It would appear that your choices are to (for an RNAV procedure) move the direct-to waypoint closer than 10 miles or (for a non-RNAV procedure) specify a radial to intercept within 10 miles.

IRL 10miles offers quite a strech to wander around if no radial or bearing is specified. You would eventually end up at the specified waypoint, but I, for example, would prolly fly quite a dogleg on the way

On a direct-to (DF), there is no track guidance. Any deviation from the line between the original initiation point and the target fix / beacon, will not be corrected for. The aircraft is just steered towards the target. This offers little obstacle protection.

Instead, a track-to (TF) or course-to (CF) will provide track guidance. If there is lateral deviation from the line defined by the track-to or course-to, the guidance system will steer the aircraft back on the original line.

Images from the FAA Traffic Flow Management learning centre


text:
Does any of the hearing protection worn by aviation professionals use active noise reduction?
I have read that passive hearing protection (standard earmuffs/plugs) offer good protection against higher frequency noise but do not protect as well at lower frequencies. Supposedly active noise reduction (ANR) can attenuate low frequency noise. Does any hearing protection worn by professionals in aviation (airports, military roles, etc.) anywhere in the world use ANR or is it all passive?

ANR can attenuate low frequency noise indeed, is mainly effective on constant or slow changing noises.

Are you asking about about pilots (headsets used in the cockpit) or ground crew?

As a backup/supplement to ANR, I make my own low frequency earplugs by wrapping 3M Push-Ins with a few inches of lead foil tape (used for ballasting golf clubs) around the shaft, covering the cylinder of tape with heat shrink.  They attenuate under 200 Hz incredibly well, at least as good as the ANR.  Light a/c are mostly 60-100 Hz.

@JohnK, color me intrigued. Do you have some pics?

@KennSebesta See answer posted below.

If you're talking about crew/pilot headsets then ANR is a very popular option - especially in General Aviation. The Bose A20 is probably the most popular headset, but both Lightspeed and David Clark also sell ANR headsets for professional aviation

My recollection is that the first Bose aviation headsets were actually the first mass produced ANR at all.  Other consumer products then followed.

Adding to the civilian answers, the F-22 helmet and F-35 helmet do include Active Noise Reduction technology. I believe most other US military fighter jets rely on passive noise reduction; this may be in part because ANR technology didn't really hit mainstream use until the 1980s, while many of the fighters in service with the US today (F-15, F-16, F-18, A-10, EA-6) began production in the 1970s or earlier.

Not quite on topic, but you if you don't have ANR units you can make very effective low frequency earplugs without spending a fortune on Flare Audio Isolate aluminum earplugs (I bought a set for 40 bucks and promptly lost them, although they do work). The metal core of the Isolates helps the foam earplug part attenuate low frequencies quite well.  If aluminum = good, lead must = better!
So I just got some 3M Push-Ins earplugs, the kind with the blue plastic shaft and a yellow bulb, some lead foil tape used for ballast on golf clubs from a golf supply store, and some quarter inch heat shrink.  If you're concerned about lead encased in heat shrink in your outer ear canal, you can also use aluminum or stainless steel foil tape for somewhat less attenuation.
Wrap a 2-3 inch length of tape around the shaft (depending how thick you want them, which depends on how wide your ear canal is), cover with heat shrink and shrink it over (I put the bulb part into a piece of metal tubing the diameter of the bulb to protect it from the heat gun).
I tested them in using stereo headphones cranked way up with low frequency sounds playing, comparing them in opposite ears against foam plugs and the Isolates (before I lost them).  These plugs attenuate the 60-70 hz noise from propeller and exhaust extremely well, better than the isolates did I would say. Above 200 Hz, they seem to be about the same as foam plugs, but the under 100 Hz noise is most of the sound in the cockpit.
I wear them under my Dave Clark ANR conversion headset (a kit from Headsets Incorporated I got to convert my 45 year old Dave Clarks) and can't really tell if the ANR is on or off.
Another tip for improving the performance of earplugs in general is to carry around a tube of lanolin (for baby bottoms) in your flight bag and apply it to the foam. They go in easier, and seal better.

I'm not sure where the actual lead in that picture so I don't know if it's valid, but my concern with this setup would be exposure to lead.

@Michael -- the lead's under the heatshrink, keeping it from contact with body fluids, or that's the idea at least

The shiny silver tape is lead tape.  It's about .010" with adhesive backing. Anyway, you're not supposed to eat it, and I normally don't have a lot of fluids oozing from my ears.  You could use aluminum or steel tape, although the attenuation you get will be lower.

Your ears constantly, continually, generate fluids.

Not oozing out.  Unless you have an infection.

I actually had a bluetooth headset once that stopped working after wearing it just due to sweat.

Well in any case, I haven't had lead oxides secreting from under the heat shrink of my earplugs over the years I've been making them.  They last for maybe 20 uses, then the foam bulbs start to go and I have to make a new set.

In case anyone wondered what's going on here: the inertia of the metal does the job at low frequencies. That's where the entire plug acts like a rigid body and just moves back and forth under sound pressure. So making it heavier does the trick. I'd personally cover that lead tape in a water-tight adhesive-lined shrink tube for extra protection, and avoid contaminating it with fingers that touched lead - they carry lots! You can use a surface lead test kit to get a sample from ear canal to test for lead. Goal is for there to be nothing detectable. A copper tape will work well instead :)

Easiest thing is to coat it with liquid electrical tape, then heat shrink over that.  Take common sense precautions, but treating it like it's uranium is maybe just a teeny tiny bit overkill.  Copper oxide is toxic as well in some dose.  So is aluminum oxide.  Plastics give off aromatic chemicals.  There's no escape.

Yes, it's extremely common, both with pilots and with passengers.
As other answers have mentioned, ANC headsets like the Bose A20 are very common with pilots. ANC is especially wonderful for light aircraft where you're sitting a few feet from the engine and often directly behind it. Light aircraft cabins don't provide nearly as much dampening/isolation of engine vibrations as something like an airliner, so the noise levels are quite high and hearing protection is a must. Even those who don't use headsets with active noise cancellation use ones with significant noise-isolating cups that fit tightly around your ears.
Aside from pilots, though, even passengers on commercial flights very frequently use ANC headphones. Many airlines provide ANC headphones to passengers in business and first class on their aircraft that are equipped for long-haul flights. Aside from ones provided by the airline, though, ANC headphones are very common with frequent flyers. The Bose and Sony ones are particularly common. I use the Bose QC35s myself when flying commercial, regardless of what cabin I'm in, as I generally find them to be better than the ANC headphones that the airlines provide. They're wonderful at reducing engine noise, even when you're sitting right beside the engine. They also help significantly for cutting down on cabin noise and are much more comfortable to wear for extended periods than headsets that have tight-fitting noise isolation cups around your ears.

Yes, they are used. You can just Google active noise-cancelling aviation headset to get a sample of many different models. I personally own a Lightspeed Zulu 3. I will also get a Bose A20 when I get back into flying.


text:
Why do big modern airplanes not use a T-tail configuration for the horizontal stabilizer?
I'm wondering why modern widebody aircraft (such as B787/777, A330/350 etc.) don't use horizontal stabilizers with T-tail configuration. What's the difference between placement of this element? There are quite a lot of military planes (C-17) and smaller ones that use it (Fokker 70, CRJ-9) in T-tail configuration, as shown here:

Related: How do conventional and T-tails differ?

Also related: Why do trijets usually have a T-tail instead of a normal tail?. Modern airliners are twins.

A horizontal tail on top of  vertical tail would require strong and heavy structures, which in turn is not good for economics of flight. The bigger the plane is, the more challenging the loads the tail has to handle are, so what might be a feasible solution on a regional jet might not work at all for a large jet.
T-tails are an obvious choice when the engines are tail mounted, and in C-17 I guess the main reason is to keep the tail away from the wake of the main wing in certain "areas" of flight envelope: C-17 is a high wing configuration, and as a military aircraft it is subjected to more "challenging" flight states.

Military cargo aircraft also need to be able to land on unimproved runways, and quickly unload large items like trucks & tanks.  The T-tail would seem to make it easier to design a nice, wide rear ramp for this.

You have good points there, the h-tail beig set higher, it is pretty much save from sand, rocks, dust and whatnot the engines and tires will kick up during ground roll/rotation. Mind if I borrow this into my answer?

@jamesqf On the other hand, there is the C130... Maybe underwing turbofans are more likely to throw up debris?

C-17  is a tactical airlifter.  That means high AOA landings at very rough improvised forward runways with lots of FOD kicked up by engine thrust.

@sdenham: The C-130 is a pretty old design (close to 70 years), and per Wikipedia inherited a lot from previous designs.  Maybe the designers back then didn't think of a T-tail, or figured the tradeoffs weren't worth it?

Must mention the British Handley Page Victor bomber, where, I daresay, economics of flight were secondary to other considerations

@jamesqf: When the torsional stiffness of the rear fuselage is compromised by a big door, a large mass on a long lever arm (= a T-tail) is the last thing you want. In order to avoid flutter, the fuselage will become heavier than with a conventional tail. Antonov understands this very well, as contrasted to Illyushin or Lockheed.

@Peter Kämpf: I'm not an aeronautical engineer by any means, but I'd suggest that observation shows that the C-17, and the C-5, which is also T-tailed, perform their jobs at least as well as their conventionally-tailed equivalents.  Any design is always going to involve tradeoffs: perhaps the rear fuselage has to be heavy anyway, in order to drive tanks on it?

@jamesqf My comment to this has been written three years ago already.

It might have something to do with the super stall phenomenon in which a stalled wing blankets a T-tail.

Isn't deep stall mostly a problem for small GA planes? An airliner could easily be designed to handle it with the autopilot, and airliners are not flown close to a stall condition anyway. The only airplanes where a deep stall is a very serious concern, are gliders, because they are flown closer to their stall conditions than any other plane. Yet even among gliders you can occasionally find T-tail configuration, and not all of them are in danger of a deep stall. For example, the IS-28b won't deep stall, yet the single-seater variant (IS-29d) can. Both have T-tails.

I should check, but iirc super stall and deep stall are not synonymous.

@vsz "An airliner could easily be designed to handle it with the autopilot". Ah yes, the 737-8 function? In short, it is a bad bad idea to design in a flight problem, that you then need to automate out for the plane to be functionally flyable. If your automation errs, your plane stops flying

@vsz Deep stall was the cause for several crashes, notably of the BAC One-Eleven in 1963 and the HS Trident three years later. Tupolev, when learning of the first crash, increased the size of the horizontal tail of the Tu-134 by 30%. The British engineers were less able to learn from their mistakes.

And as for super stall, a good example would be the saab draken, with it's hideous stall characteristics...


text:
Zenith-Add horsepower, NASA-add drag: to create lift. How is this possible?
Available horsepower and adding drag to an airplane both create extra lift?  How is that possible?

Available Horsepower
According to the Zenith, available horsepower not used to overcome form drag and induced drag contributes to rate of climb.
Adding drag
NASA's report on the X57, states that drag might need to be added to get rid of excess thrust to create more lift.
From the conclusion of that NASA report on span-wise propellers generating lift:

A potential constraint on the design of the high-lift propellers system is the total thrust. If large amounts of lift augmentation are required from the propellers, then high thrust values from the high-lift propellers are likely. If these propellers produce excess thrust, then the aircraft will be unable to sustain flight at the desired speed. This implies that drag producing devices may need to be added to the aircraft or the primary propellers operated as windmills to produce additional drag.

(High-Lift Propeller System Configuration Selection for NASA's SCEPTOR Distributed Electric Propulsion Flight Demonstrator, NASA)
This concept seems to be exactly opposite of the first concept!!
Rumours: Ultralights reach a max speed, any extra power is then converted to lift. Is this true?

How do these concepts work?

Do you have a direct quote from the NASA report to support #2?  Because you are right, it makes no sense.  Context is needed to determine if you misinterpreted it somehow.

Source and direct quote from NASA report added above.

@Fred, based on the quote you provided your interpretation that adding drag will create lift is incorrect. Basically what it is saying is that if you rely heavily on propellers to create lift, you may end up with a lot more thrust than you need for normal forward flight. That's really all it's saying. Adding drag to compensate would be silly.  As far as # 1 & 3, if you are overpowered you can either speed up, pull back the throttle to slow down, or climb. What specifically about the concept of excess power do you not understand?

You are mixing a lot of concepts together. First you don't need a lot of thrust to climb. To climb fast you need fly steeper which means your engine needs to fight gravity harder in order to not slow down. You actually need less lift in a climb than flying level. Lift argumentation is a totally different concept that I has nothing to do with 99% of production airplanes. In this case it's only meaningful within the context of that paper and that plane.

@Michael Hall: I think (based just on the quote) that the problem is that it can't fly slowly enough.  If the propellors produce enough lift, they also produce thrust which makes the plane go fast.  It'd be something like trying to land your high-performance sailplane without spoilers or speed brakes.

@jamesqf, isn't that the same thing I said?  ("you may end up with a lot more thrust than you need for normal forward flight")  And if I wasn't clear, adding temporary drag for landing in the form of speedbrakes is not silly, it would only be silly as a permanent design solution.

@user3528438, please explain your comment:  "You actually need less lift in a climb than flying level."

He probably meant flying at top speed rather than a slow economy cruise climb

Re Item 2:  NASA should have stated: drag devices may be required to fly at slower landing speeds.  The concept of speed brakes is nothing new, whatsoever.

@Fred - "He probably meant flying at top speed rather than a slow economy cruise climb."  It doesn't matter.  In steady state lift must always equal weight regardless of airspeed, or whether you are flying straight and level or in a constant rate climb or descent.

@Michael Hall: Communication problems :-(  It may be what you intended to say, but that wasn't what I understood you to say.

Equating rate of climb with more lift is wrong. In order to climb the aircraft needs more thrust but a bit less lift. Therefore, the first statement is wrong: Adding more power does not create more lift. The only case where this happens is at high speed and the maximum sustainable lift coefficient: More power allows to fly at a higher load factor, but depending on the flight envelope, might result in a crash. The explanation does not make that connection, so it is wrong.
The second explanation is special for the X-57 because the concept uses the propeller slipstream to create blown lift. At low speed, the thrust needed to maintain flight speed is low and might not be sufficient to create enough lift on the small wing. Adding some drag allows to fly at higher thrust without climbing away. But only in this special case when most of the wing is in the propeller slipstream. So the second statement is correct but only for a single case.

Artist concept of the NASA X-57 Maxwell (image source).

If I increase throttle, it adds horsepower, which allows an increases in engine rpm, which increases thrust, which increases speed, which causes me to climb because the wings are producing more lift, no?

How does distributed propulsion (DP) work in a pusher configuration, if the props don't add airflow to the wings?

Re my comment above, I'm talking about climbing in straight and level flight, not pitching the aircraft up. eg. economy cruise climb at 100ft/min

@Fred: If you add power you increase the flight path angle. Angle of attack will be unaffected, details depending on the power impact on the pitch equilibrium. Climb has noting to do with more lift, only the initial acceleration to a higher climb speed needs a bit of extra lift. There is no "straight and level" climb. Either you climb or you fly level. Please read the linked answer!!!

@Fred Even pusher props will add a bit of extra flow speed over the wing, but a lot less than tractor props. Basically, distributed propulsion with pushers is possible but has little impact on lift.


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What is the total EXTRA lift produced using an 80 deg flap and distributed propulsion?
What is the total EXTRA lift produced using an 80 deg flap? ( 50 deg flap at 30 deg AOA ) and distributed propulsion.
Specs: Ultralight, stall speed 20mph, ROW of 12 x 15" props along entire wing span in DISTRIBUTED PROPULSION, max local AOA estimated at < 9 deg ), electric 25 hp (total), MTOW 500 lbs, low wing in ground effect, max 3 ft height for xSTOL/nVTOL, VG's, full span single slotted flaps with spoilers , Clmax: 3.4.
Hypothesis:
A low wing ultralight accelerates with a short field takeoff, along a runway to 20mph and rotates to 30 deg AoA just prior to takeoff, producing the following:
Total lift = Wing lift (bernoulli/momentum/vortex lift + deflection lift )  + Prop lift + Flap Lift
Example:

Wing lift:

standard wing lift formulas, lift equal to airplane weight of 500 lbs.

Prop lift:

Estimated thrust at 20mph: 5lbs/Hp thrust x 25 HP = 125lbs thrust.
Prop lift = 125 lbs x sin ( radians(30 deg))= 62 lbs

80 deg Flap Lift

Thrust due to Prop wash on bottom of wing hitting flap
= 30%(???) of total thrust = 37 lbs.

Total extra vertical thrust/lift

= 62 + 37 = approx 100 lbs ( 20% extra vertical thrust/lift...!!!! )

Reduced MTOW wing requirements at Vtakeoff to Vflap

orginal MTOW = 500 lbs.
MTOW revised= MTOW original - extra vertical lift = 500lbs - 100 lbs.= 400 lbs.
This is the reason for asking the question:
Could I then resize my wings based on 400 lbs MTOW at Vtakeoff(Vso) to Vf, not 500 lbs?
Is this correct thinking?
See black tuft dwg below showing downwards air flow.

Thrust is generated by the engine. Did you mean lift or drag?

The edit makes a little more sense. But, are we to assume the engine is directing its slipstream against the underside of the wing. Otherwise, the only contribution to lift is made by the relative wind, not by the engine. Am I missing something?

Maybe I am thinking about this wrong. Help me with the following scenario:

You are idling at the AER in a pusher-prop ultralight with the brakes on. Thrust and Lift are negligible. You go full throttle. Thrust increases to X but Lift remains only a small percentage of X. You release the brakes. Lift increases as airspeed increases. Thrust remains X. After reaching TPA, you chop the throttle to idle. Thrust almost immediately becomes negligible. On the other hand, Lift declines in proportion to airspeed. Am I thinking about this wrong? How are you generating Vertical Lift with Horizontal Thrust except through relative motion through the airmass?

Otherwise, only the lift that is deflected by the flaps from a tractor-prop is converted to Vertical Thrust. The rest remains as slipstream to create Vertical Lift. That begs the question of how many ultralights have flaps positioned behind the prop, in its slipstream?

Vertical Lift is thrust.  A propeller is a little wing going in a circle.  A helicopter rotor is a longer wing going in a circle.  A regular wing goes straight.  It's all the same, except for circle/straight part.  Fred seems to be asking what is the lift increase for the main wing as a result of the propeller slipstream - the circular wing's horizontal "downwash" - impinging on the flap.

@JohnK - Although Thrust from a prop is a form of Horizontal Lift, I don’t think that is what Fred is implying. And, the question has nothing to do with rotorcraft’s Vertical Thrust/Lift.  It is, as you said, about the increase in Lift due to prop slipstream. But, Fred goes a step further. Asking how much lift is increased even more when lowering flaps to 80°. The 20 knot relative wind should not be taken into consideration for this question. Only the force of whatever slipstream contacts the flap. And, how many ultralights have flaps positioned behind the prop, in its slipstream?

The most recent edits make much more sense than the original question. Especially since you have added that you are considering a row of props along the leading edge. Try researching the following aircraft for a lead on your answer:  https://www.tecnam.com/us/innovation/p2006t-x-57-maxwell-nasa/

A 30° flap is a typical landing setting. Increasing that to 80° is only adding more drag. Your 25 hp thrust will only add a mild breeze and will not help to reach that 30° AoA unstalled. Better use 8° or 10° of flaps (depends on flap chord) and do not rotate to more than 10° AoA.


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Can endplates be used to reduce chord length by up to 14%?
Can endplates be used to reduce chord length by up to 14%?
I understand large endplates can change the lift distribution from an elliptical one to a uniform one, making the wing more efficient by increasing lift near the wing tips.
This increases the effective aerodynamic span by up to 16% ( see pic below ).
pic: Effect of large endplates on a short span wing, Source: Zenith

Can this be used to shorten the chord?
eg.
16' span, 2' chord, total wing area: 32 sq ft rectangular wing.
Use large end plate, h/b=20%.
revised effective aerodynamic wing span: 16' x 1.16= 18.5'
Revised chord: 32 sq ft / 18.5 = 1.7' chord.
Is this correct thinking?

I read somewhere many years ago that endplates have to extend about a full chord length beyond the airfoil profile to do much good, at least a net benefit over their own drag.  Which is why you rarely see them.  The ones you sometimes see on STOL modded airplanes that stick up a couple inches are useless.

Why would chord be any different? Just because the effective span is increased, doesn't mean the physical area has increased.

It would be possible to do as you suggest.
However the short-chord wing would have to be either thinner and therefore heavier, or less streamlined and therefore less efficient causing higher form drag. Both effects increase induced drag.
The endplate would need to be roughly square in proportion, which would add drag of its own. Also its mass and side area would complicate the aircraft behaviour.
It is generally better to add a small winglet rather than a big flat plate and use it to reduce the wing tip vortex. But even there, a bit extra wing span with suitable washout is just as good.

But can one use the longer aerodynamic span instead of shorter physical span for chord calculations?

I would not muck about with correction factors, they are really only advisable for sanity checks. Large endplates will allow a greater total lift because it does not taper off, so the flatter lift profile will allow a higher mean wing loading. Once you decide on the weight, span and wing loading then the area and mean chord will fall out of that.


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Why Do General Aviation Aircraft Cost So Much?
Purchase price. Not total cost of ownership.
This Aviation Question inspired mine.
A quote From this article in Flying Magazine (Dated 2012):

Cessna is holding the line on pricing, but the new models still aren’t cheap. The 180 hp 172S, referred to by Cessna as the 172-SP goes for \$307,500 the 160 hp 172R sells for \$274,900.

Many years ago I heard this generalization: "half the cost of an airplane is litigation." And sure, I understand you should not replace a control surface cable with one from Home Depot. But still, these prices are breathtaking.

Yeah, this is probably a dupe. Question title words just didn't get that hit I guess

Just as an FYI, club/group ownership is another option to look at. Really spreads out the ownership costs and makes things affordable. I bought my "share" of a very well equipped, 1978 Cessna 177B Cardinal II for less than $8k. My cost per hour (wet) is less than the local school charges for its 152, and the airplane see's maybe 50 hours a year by all members.

In reading the earlier answer: I get a sense of a self fulfilling prophecy. Low volume because they cost too much. Then there's Less mgr. liability now. Then How many times does a 50 yr old design get certified anyway? Upgraded radio? How much certitude does that take? There's more gone into my car that a C-172 I think. Its not rocket surgery.

A lot of it is self-fulfilling prophecy. Also bear in mind that for your typical light GA single (say the Cessna 172 from your example) the purchase audience isn't generally individuals: They're selling these things to major flight training programs - often for much less than the "sticker price". Straight-line depreciation on a working airplane & the money they'll make when they sell it off in their next fleet refresh help soften the blow.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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What is this device mounted on the fuselage of Lufthansa planes?
I noted the device shown in the photo mounted on top of the fuselage near the tail. The photo shows a smaller airbus (I think) but it is also present on several A380ies, and a Boeing 747. In fact, most, but not all Lufthansa planes (observed at Frankfurt airport) seem to have this but I haven't noted it yet with other airlines.

A satellite antenna for passenger internet. See https://aviation.stackexchange.com/questions/14849/how-does-wifi-work-on-an-aircraft/14850#14850 and https://aviation.stackexchange.com/questions/46544/how-does-an-airbus-aircraft-connect-to-the-internet/46547#46547

Thanks, that solves it (although without your comment, I could not have found the duplicate question) - my sample of non-Lufthansa planes was obviously too small.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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What is meant by a “cold soak" of an APU?
What is the exact meaning of "cold soak" of an Auxiliary Power Unit (APU)? Is it to do with long flights at high altitude and the fuel getting cold? Will it affect the ignition of the APU? Can anyone tell how this affects the function of APU?

It means the APU has been in active or not operating in a cold environment for an extended period of time. Here, like any other piece of machinery and operating fluids are “cold soaked“, that is, has reached the ambient temperature of the environment.   During extended flight at high-altitude, and APU would encounter this as it is not normally used during cruise flight.   This does affect the ability of the unit to power up in the event of an emergency, as most machines don’t like being operated in extremely cold weather, and usually require a period of time in order to warm up.

Whenever a product go through HALT/HASS testing there it will be socked in different temp ranges and even on the pressure as well. As its already handled in that testing why we are doing it again? When we do this kind of testing the product life will be reduced >50% (my college told me this) so why is it done again? or am i getting i wrong

The APU "cold soak" in flight start program is part of the ETOPS airworthiness requirements.
The ETOPS Reliability program requires that in twin engines aircraft to periodically Check the ability of the APU to start in flight when "cold soak", it means the temperature has decrease as much that the fluids and mechanical components become freeze.
To be representative, the "cold soak" can be achieved in flight after 2 hours of the APU shutdown and before the top of decent.
You can find detail information in FAA AC 120-42, par.301m.

The term 'cold soak' is used, for example, when the aircraft is parked overnight in low temperature weather conditions.
It is possible that the FADEC APU presents issues igniting the APU due to fuel low temperature.
For more info on aircraft cold soak tests, check the following videos:

Welcome to Aviation.SE!

Whenever a product go through HALT/HASS testing there it will be socked in different temp ranges and even on the pressure as well.
As its already handled in that testing why we are doing it again?
When we do this kind of testing the product life will be reduced >50% (my college told me this)  so why they do many times? or am i getting i wrong


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Why is the landing distance increased on grass runways?
Checking the Cessna 152 POH it says to increase takeoff distance by 15% of ground roll (makes sense due to increased friction). Also says to increase landing distance by 45% (doesn't seem to make sense). Would like to understand why.

It says increase landing distance by 45% in wet grass I believe.

And takeoff distance too... The most scary takeoff I remember was on a grass surface, long grass and very wet with the dew of the early morning. I thought that the runway available was enough, and that 'that of the wet grass' was just something 'in the books'... To this day, I  believe that I survived that takeoff thanks to some lift from Above...

The long and short of it is that wheels are designed for smooth tarmac. While grass increases friction on the wheels on takeoff (making the aircraft work harder to get up to speed), it also reduces the efficiency of the brakes, thus taking longer to slow down.
And to add to @GdD's point, grass retains moisture much better than tarmac, so in wet conditions it is considerably slippier (and will remain so for longer afterwards).

O.K. thanks. I normally don't use the brakes very much or hard so I didn't factor this into the equation. 45% can be quite significant.

@Larry Fair enough, it's worth remembering that grass runways can (and often are) shorter than the runway you might be used to, so don't scrimp on those brakes!

I have to disagree: wheels are NOT designed for smooth tarmac.  They are designed to work on any fairly smooth surface, whether it's tarmac, grass, concrete, dirt, or (in northern parts) a frozen lake.  It's just that those different materials have different properties, which change the amount of rolling friction (or sliding, if you brake too hard).  Same principle would apply if you're landing on say a paved runway with patches of snow or ice.

@jamesqf They may be designed around a multitude of surfaces, but unless you've got very specialised gear the best performance will still be found on paved runways.

@Cooper: Sure, that's why POHs recommends increasing takeoff & landing distances on different surfaces.  You probably learned something about this in basic flight training, too, just as you learned to take density altitude into account.  (I'd expect the reason the base distances are quoted for pavement is simply that it's repeatable, while grass & dirt vary a lot.)

On the other hand, aren't grass runways generally much softer than paved runways (unless you've come up with a strain of grass that grows on concrete), which one would expect to decrease the landing roll needed (though further increase the takeoff roll)?

@Sean Without using any brakes I suppose, yes.


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Are there FAA rules about changing an airport name from Municipal to Regional?
Are there any FAA requirements to be met or stipulations for an airport to change its name from Municipal to Regional?

In general there are no strict rules on airport names. This would be more of a marketing decision, and it's up to the owners/operators to decide of the expense of the "rebranding" is worth the benefits.
US regulations do cover international airports, and the Secretary of the Treasury designates the official list of airports. Even then, there are plenty of international airports not on this list, and there are even some airports called "international" with no customs facilities or plans to have them in the near future.

The FAA does outline what a regional airport is here

Regional:
Supports regional economies by connecting communities to statewide and
  interstate markets.

There is also a lot of interesting info on regional airports here. However neither of those documents cover the "Municipal" designation.

Those definition seem to be more in support of the study to guide regulations and funding, not really an impact on naming. Their example of a regional airport is named a municipal airport.


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Which prop location creates more lift? Leading edge or blown flap?
Which prop creates more lift? Leading edge or blown flap?
In distributed propulsion there seem to be 3 prop locations: leading edge (NASA X57, blackfly), blown flap (Lillium), and pusher prop (Kittyhawk).
Which would create more lift, leading edge or blown flap?

Leading edge: The entire wing chord sees about 50% of prop exit speed (turbulent though), the prop pushing 50% of the air under the wing (although it would create lift by hitting the flap (assume 45-90 deg flap position)
Blown Flap: wing sees 100% of prop exit speed but only on about 45% of the of the chord ( assuming 50% chord flap), but the first 45% of the wing sees non turbulent air. Blown flap has the advantage of providing vertical thrust.

See pictures below.
Lillium - blown flap ( assume props, not the actual electric turbines it uses)

Kittyhawk - pusher prop

Nasa X57 - leading edge/tractor prop

Below are 3 Cp plots from xflr.
First plot has no flap, but has a roughly 30 deg angle in (-) pressure drop.

Adding a 50% chord flap, increases the suction at 0% chord???
Was surprised to a spike in negative pressure at 0% chord from
a 50% chord, 45 deg flap!!! Would have thought there would be a negative pressure spike at 50-60% chord....

What would the shape of the pressure plot be when distributed propulsion is added to the flap? See plot below.

Your airfoil's upper side seems to have separated flow from 40% downstream. The 50% flap stalls the airfoil.

Why doesn't NASA just build that cute sporty with 2 large wing tip props and slats and flaps???

@RobertDiGiovanni: Because they first need FFA approval. With empty batteries it will be incapable of 61 kts minimum speed. If you just want the cute plane (without tip props), get the original from Tecnam.

@Peter Kampf well, it is an electric design, but may be capable of lower glide speed with slats and flaps.  Also, I would look for ways to avoid empty battery scenario, such as a backup generator.  I agree 2 large props would produce a lot of drag...unless they were feathered!  Good link to the original Tecnam.

Why is an empty battery an issue? Isn't that just like running out of fuel? Wouldn't a reserve quantity like in in gas engines just solve the problem?

@Fred: The design relies on extra speed over the wing for lift at minimum speed. No power, and the X-57 stalls at maybe 70 knots. That is too much for a part 23 certification.

The X-57 should show the highest gain in lift from the props. They will wrap the full wing in accelerated flow. XFLR is not capable of capturing this effect, since the pressure distribution should be similar to the case without props, only the reference dynamic pressure is higher than that of the prop-less wing. Of course, swirl effects will add a spanwise pressure variation that is missing without the props.
Next best is the pusher arrangement of Kitty Hawk. Here the props can swivel, so maximum lift is less of a concern. The prop arrangement has been chosen to minimize noise from the prop blades cutting through the wing wake; that should explain the considerable distance between props and wing. At that distance the suction of the prop will affect the flow over the wing much less that what the X-57 should be capable of. In a pressure plot the main effect should be a delayed flow separation: The wing will tolerate a slightly higher angle of attack.
Worst of all is Lilium: Here the inefficient, small-diameter props sit in ducts which all have their own boundary layer and losses. Next, any accelerated flow only affects the upper side of the wing, so half of the effect will be lost. This arrangement, however, will make the plane sound much louder and powerful than Kitty Hawk's. Judge for yourself whether that is a worthwhile achievement.

Lilium is showing a much higher power requirement to lift and cruise 5 people.  Helicopter data could be in this study (which features one very large (more efficient) rotor).

What is "half the effect will be lost"? Is it the prop wash hitting the flaps, generating a reaction force upwards?

@Fred: The accelerated flow only works on the upper wing of Lilium. Compare with the two other concepts where it affects both sides.

I guess I don't understand how accelerated flow on the bottom of a wing creates upwards lift except for the reaction force of the upward swirl hitting the bottom of the wing or air hitting the flaps and deflecting downwards. I thought it is mostly the camber of the upper surface that creates lift, no?

@Fred: Lift can be explained in several ways. It is the reaction force to the downward acceleration of air flowing over the wing, which in turn needs a pressure difference to work. When more air is involved, the forces grow. If only one side has that benefit, only one side (or half the possible airflow) benefits.

Re Lillium: I thought ducted fans are slightly more efficient (16%??) than props? Also, wouldn't the Lillium have a much higher prop exit speed with small props, and therefore a high gain in lift as the wing would see possibly 200%+ increase in speed in STOL ( but terrible efficiency in cruise)?

Re Kittyhawk: I don't see how the Kittyhawk would see any gain in lift as the wing would see zero increase in free velocity as the props are so far behind the wing.

@Fred: If you compare props and ducted fans of the same diameter, maybe the ducted fans are more efficient at some operating point. If you are free to choose the diameter, the prop wins hands down. It's not even funny - the ducted fan is left in the dust. The prop distance of the Kittyhawk certainly reduces their effect on lift, but I would not be so sure that it is zero. But it will not be large, either.


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Is there any benefit to sweep back on an ultralight?
Is there any benefit to sweep back on an ultralight of:

15 deg
30 deg
45 deg.
other?

There are two primary benefits of wing sweep at ultralight speeds. The first is that it acts as variable dihedral -- the higher your angle of attack, the more dihedral effect you get from sweep.
Secondly, when you see ultralights (or hang gliders) with a lot of sweep, it's generally for a tailless design that needs to get the tip fins/rudders and twisted tips and/or elevons back behind the CG location so they have a useful moment arm.
The benefits of sweep in transonic flight (delay of Mach effects onset) obviously don't apply at ultralight aircraft speeds.

You sometimes end up with too much dihedral effect and are forced to build in geometric anhedral, like the Swift ultralight glider, and most high wing swept wing transports.

Possible restructuring: put second paragraph first, deleting first 3 words.  Put first paragraph second, dropping first 8 words and saying "Also,  sweep acts as a variable dihedral-- the higher your angle of attack..." Add second sentence to this paragraph saying "This may or may not be beneficial. Many ultralights actually have some anhedral to counteract this effect." Motivation for restructuring: in practice, wing sweep in ultralights is usually there for the benefit of the pitch axis dynamics, not for the the purpose of increasing the dihedral effect. Something like that, you get the idea.

For an ultralight aircraft with a wing and tail design and a top speed of less than 60 mph, sweep of greater than 15 degrees does not serve a useful purpose.
However, some sweep can help solve a discovered center of gravity issue by moving the center of lift rearwards.  This was commonly done on bi-planes such as the Tiger Moth well before the supersonic age.
Another benefit of sweep is increased yaw stability.  This is not to be confused with the Dutch Rolling nightmare of large aircraft.  A light aircraft will benefit from the stability if needed.  Interestingly, this means a smaller vertical stabilizer but a larger rudder!
Forward sweep has also been tried in gliders to make them more responsive to rudder inputs.
Sweep also increases the cross-wind "dihedral effect", which rolls the plane away from the crosswind component.  For highwings, this can be used to balance the "anhedral effect" of side area under the CG, creating more desirable cross wind handling characteristics.  The Cessna 172 is a fine example of near perfect cross wind effect balance.
But sweep can be detrimental to the stalling characteristics of an aircraft, as swept wings tend to start stalling near their tips.  Washout or leading edge slats help prevent this.  So, it is to be approached with caution, and full understanding of its effects.  7 degrees would be more like it, for starters.
The Electric Wren motorglider from the 1920s is a great ultralight reference from the days when slow flight was thoroughly understood and the pilots were glad of it.
Although highly swept and delta type designs were considered for this answer, in slow flight straight wings are more efficient, requiring less power, there for a lighter engine. The Electric Wren flew on less than 10 horsepower!


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Would a 2nd row of distributed props on a 2nd staggered wing of a biplane increase its efficiency?
Would a 2nd row of distributed props on the rearward wing of a biplane with 25% gap and 150% negative stagger increase the efficiency of a biplane ULTRALIGHT, 25MPH, Re 500K? ( rear wing is 25% lower than forward wing, 150% rearward)
Both wings would have distributed propulsion from several props along their span in tractor configuration.
Forward wing props see a prop inlet speed of V1, or aircraft speed.
Forward wing sees V2 ( exit speed of prop), (assume an increase in speed of 50%, based on prop size, Hp using prop formula) V2=150%V1
Calculate forward wing size based on 150%*V1*k1
k1= efficiency/coverage factor of Wing1
Props of 2nd (rear) wing would have a inlet speed of V2, but an exit speed of V3 ( assume another 50% increase)
2nd wing sees a speed of V3=150%V2=150%(150%V1)=225%V1.
Calculate rear wing size based on 225%*V1*k2
k2=efficiency/coverage factor of Wing2
An interesting note: average leading edge airspeed over wings is (150%+225%)/2=190% of aircraft speed!! ( if this proposed theory is correct)
Is this correct, at least conceptually? See side view of configuration below.


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Why does an aircraft's nose tend to pitch downward at higher bank angles?
In the usual aircraft configuration, the center of lift is located behind the CG producing negative pitch moment (the horizontal stabilizer counteracts it by producing downforce - positive pitching moment). If the aircraft rolls to one side the vertical component of lift on the wings would decrease, so by that logic the aircraft should pitch up.
Why does it pitch down? What am I missing?

Notice at higher bank angles rudder into turn pitches nose down more and more as well.  Very important to stay well above higher stall speeds in steep turns, also helps to relax rudder a bit and let adverse yaw hold nose up.  At very high angles, opposite rudder may be needed.  But this is advanced and should not diminish in any way the importance of coordinated turns for safety, especially near stall.

Perhaps it would be more pertinent to have asked this question as, "Why does the nose get lower in relation to the horizon at higher bank angles without increased pitch."  The question disregards the point that the axes are no longer in the same position relative to level flight once in the turn.

"If the aircraft rolls to one side the vertical component of lift on the wings would decrease" -- why?  This question appears to be built around a flawed premise.

The answer is that the aircraft doesn't actually pitch down. When banking (or rolling) into a turn, the angle of attack (or the angle between the center of the wing and the flight path of the aircraft) stays the same. Assuming no correction to maintain altitude is made, the aircraft will continue on the same angle of attack, just banked over by some amount. Due to the bank, the lift produced by the wing is no longer vertical, instead, it is directly perpendicular (at least for our purposes) to the wings. Therefore, if the airplane is in a 30 degree bank, then the lift vector will also be 30 degrees from vertical. Doing some vector analysis, then you can determine that if the angle of attack hasn't changed the new vertical component of lift will be about 86% of what it was in level flight. Therefore, the plane would begin losing altitude. This altitude loss is what you see as the airplane pitching down.

Of course the aircraft does actually pitch down. What you described is the first part, but then the longitudinal stability kicks in, the angle of attack will increase as the flight path angle turns down, but pitch does not, it will cause reduction of the down-lift on the elevator (and increase of lift on the main wing) and thus pitch-down moment.

@JanHudec Exactly what I was looking for, thanks. In a turn AOA gets little higher -> increases lift on the wings and decreases downforce produced by the tailplane -> negative pitch moment and nose drops. I would mark yours as an accepted answer but it is a comment..

@Darjan this doesn't happen. The downforce produced by the tailplane is never reduced! It is the lack of lift from the main wing (because of the non-vertical lift vector) that leads to the nose drooping. The tail in fact then must create MORE downforce when the pilot pulls back on the stick to maintain altitude.

@PilotDan, if the pilot changes the control input, what the aircraft does obviously depends on that change. If the pilot pulls back on the stick enough to maintain altitude, the aircraft raises the nose, because it needs higher angle of attack now. However, usual understanding of “tends to pitch” is “in absence of control input”—and in that case the sequence is bank decreases vertical lift, aircraft accelerates downward (without pitching yet), this increases angle of attack and that finally pitches the aircraft down due to decrease of tail downforce along with increase of wing lift.

But tail cannot produce pitch down moment as well. Now tail has to push agains centrifugal force. But gravity is still pulling nose if the flight is not coordinated so it will go down.

Darjan,  first the basics.  An aircraft in straight and level flight has the center of all lifts balanced with the center of gravity.  A plane with the center of gravity set forward of the WING center of lift requires down force on the tail to balance it.
Now roll 45 degrees.  Gravity vector is no longer aligned with elevator trim.  If you roll 90 degrees the rudder is now you "elevator" relative to CG.  Unless you apply opposite rudder, a
forward CG set nose will drop.  This, among other reasons, is why aerobatic planes like their
CG further back.
Agree with @PilotDan that bank will also cause reduced vertical lift, but a forward set CG will also drop the nose.  Many pilots will try to pull harder on the yoke to maintain altitude, adding power seemed to work better for steep turns, as well as reducing bank angle.
Bottom line, forward set CG is not a panacea of safety in all cases.  Published forward CG
limit is best adhered too.

Really would appreciate a comment with downvote.  This is about the 5th one in the last 2 days.  Let's please try to be constructive here and say why. What is your opinion?


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Why does strong shock wave-boundary-layer interaction produce a bubble reattachment?

The picture represents a lambda type shock wave, separating the boundary layer at its foot. Separation causes a recirculation bubble to appear. I understand why separation has happened, but what is the exact reason for a bubble reattachment?
The picture is from the book called Shock Wave-Boundary-Layer Interaction by Holger Babinsky and John K. Harvey, Cambridge University.

After the rear shock leg, the subsonic flow inside the boundary layer decelerates further, which causes an expansion of the streamtubes and therefore reattachment. Note that with the presence of the supersonic tongue, the deceleration into the subsonic regime happens somewhat "slower" inside the boundary layer than outside (especially above the slip line).